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Asymmetric Simple Exclusion Process with Open Boundaries and Quadratic Harnesses

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Abstract

We show that the joint probability generating function of the stationary measure of a finite state asymmetric exclusion process with open boundaries can be expressed in terms of joint moments of Markov processes called quadratic harnesses. We use our representation to prove the large deviations principle for the total number of particles in the system. We use the generator of the Markov process to show how explicit formulas for the average occupancy of a site arise for special choices of parameters. We also give similar representations for limits of stationary measures as the number of sites tends to infinity.

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Acknowledgements

JW research was supported in part by Grant 2016/21/B/ST1/00005 of the National Science Center, Poland. WB research was supported in part by the Taft Research Center at the University of Cincinnati. We thank Amir Dembo for a helpful discussion and references on the semi-infinite ASEP.

Author information

Authors and Affiliations

  1. Department of Mathematics, University of Cincinnati, PO Box 210025, Cincinnati, OH, 45221–0025, USA

    Włodek Bryc

  2. Faculty of Mathematics and Information Science, Warsaw University of Technology, pl. Politechniki 1 00-661, Warszawa, Poland

    Jacek Wesołowski

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  1. Włodek Bryc
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Correspondence to Włodek Bryc.

Appendix: Proof of Theorem 3

Appendix: Proof of Theorem 3

Denote \(\left[ n\right] _{q}=1+q+\ldots +q^{n-1}\). Consider the family of monic polynomials \(\{Q_n(y;x,t,s): n=0,1,\ldots \}\) in variable y, defined by the three step recurrence:

$$\begin{aligned} y\ Q_n(y;x,t,s)=Q_{n+1}(y;x,t,s)+{\mathcal A}_n(x,t,s)Q_n(y;x,t,s)+{\mathcal B}_n(x,t,s)Q_{n-1}(y;x,t,s), \end{aligned}$$
(6.1)

with initial polynomials \(Q_{-1}\equiv 0\), \(Q_0\equiv 1\), where the coefficients in the recurrence are

$$\begin{aligned} {\mathcal A}_n(x,t,s)=q^n x+\left[ n\right] _{q}\left( t\eta +\theta -\left[ 2\right] _{q}q^{n-1}s\eta \right) , \end{aligned}$$
$$\begin{aligned} {\mathcal B}_n(x,t,s)=\left[ n\right] _{q}\left( t-sq^{n-1}\right) \left\{ 1+\eta x q^{n-1}+\left[ n-1\right] _{q}\eta \left( \theta -s\eta q^{n-1}\right) \right\} , \end{aligned}$$

for \(n\ge 0\). (For \(n=0\), the formulas should be interpreted as \({\mathcal A}_0(x,t,s)=x\), \({\mathcal B}_0(x,t,s)=0\).)

It is known, see [9], that family \(\{Q_n(y;x,t,s): n=0,1,\ldots \}\) is orthogonal with respect to the transition probabilities \(\{P_{s,t}(x,dy)\}\) of the bi-q-Poisson process, i.e.

$$\begin{aligned} \int Q_n(y;x,t,s)Q_m(y;x,t,s)P_{s,t}(x,dy) =0 \text{ for } m\ne n. \end{aligned}$$
(6.2)

Polynomials \(M_n(y;t):=Q_n(y;0,t,0)\) are martingale polynomials, i.e.,

$$\begin{aligned} \int M_n(y;t)P_{s,t}(x,dy)=M_n(x;s) \end{aligned}$$
(6.3)

and (6.1) simplifies to the three step recurrence:

$$\begin{aligned} x M_n(x;t) =M_{n+1}(x;t)+(\theta +t\eta )[n]_qM_n(x;t)+t(1+\eta \theta [n-1]_q)[n]_qM_{n-1}(x;t), \end{aligned}$$
(6.4)

\(n\ge 0\), with \(M_{-1}=0,M_0=1\). In particular, it is clear that \(M_n(x;t)\) is a polynomial in both x and t.

In view of (6.3), the action of the infinitesimal generator on \(M_n(y;t)\) is simply

$$\begin{aligned} A_t(M_n(\cdot ;t))(x)=-\frac{\partial }{\partial t}M_n(x;t); \end{aligned}$$

this is easiest seen by looking at the left-generator, compare [13, Lemma 2.1]. Note that by linearity this determines action of \(A_t\) on all polynomials: for \(p(x)=\sum _{k} a_k(t)M_k(x;t)\), we have

$$\begin{aligned} A_t(p)(x)=-\sum _k a_k(t) \frac{\partial }{\partial t}M_k(x;t). \end{aligned}$$
(6.5)

It is enough to determine action on polynomials of the related operator \(H_t(p)(x):=A_t(x p(x))-x A_t(p)(x)\), compare [13, Eqtn. (13)], and it is enough to determine action of \(H_t\) on polynomials \(M_n(x;t)\).

Lemma 2

$$\begin{aligned} H_t(M_n(\cdot ;t))(x)=\eta \left[ n\right] _{q} M_n(x;t)+(1+\eta \theta \left[ n-1\right] _{q})\left[ n\right] _{q}M_{n-1}(x;t). \end{aligned}$$
(6.6)

Proof

This proof is a minor variation of the proof of [13, Lemma 2.2]: we use (6.4) to write \(x M_n(x;t)\) as a linear combination of \(\{M_k(x;t)\}\), apply (6.5), and then subtract

$$\begin{aligned} x A_t(M_n(\cdot ;t))(x)=-x\frac{\partial }{\partial t}M_n(x;t)=-\frac{\partial }{\partial t}\left( xM_n(x;t)\right) , \end{aligned}$$

where we again use recursion (6.4). \(\square \)

Lemma 3

If p is a polynomial, then

$$\begin{aligned} H_t(p)(x)=(1+\eta x) \int \frac{p(y)-p(x)}{y-x} P_{q^2 t, t}(q(x-t\eta )+\theta ,dy). \end{aligned}$$
(6.7)

Proof

Fix \(t>0\). The first step is to go back to recursion (6.1), and notice that \(\nu _{x,t}(dy):=P_{q^2 t, t}(q(x-t\eta )+\theta ,dy)\) is the orthogonality measure of the polynomials \(W_n(y;x,t)=Q_{n+1}(y;x,t,t)/(y-x)\), \(n=0,1,\ldots \). This is because polynomials \(\{W_n(y;x,t)\}\) satisfy the three step recursion

$$\begin{aligned} y W_n(y;x,t)=W_{n+1}(y;x,t)+ \left( q^{n+1}x+\left[ n+1\right] _{q}(t\eta +\theta -\left[ 2\right] _{q}q^n\eta t)\right) W_n(y;x,t) \nonumber \\ + t\left[ n+1\right] _{q}(1-q^n)\left( 1+\eta x q^n +\left[ n\right] _{q}\eta (\theta -t\eta q^n)\right) W_{n-1}(y;x,t),\nonumber \\ \end{aligned}$$
(6.8)

which is derived from (6.1) with \(s=t\).

Clearly, (6.7) holds for a constant polynomial. By linearity, it is enough to verify (6.7) for \(p(x)=M_n(x;t)\) with \(n\ge 1\), in which case the left hand side is given by (6.6). We want to show that the right hand side is given by the same expression.

For a fixed \(s<t\) and \(n\ge 1\), we write polynomial \(M_n(y;t)\) as a linear combination of the (monic) polynomials \(\{Q_k(y;x,t,s):k\ge 0\}\) with coefficients \(\{b_{n,k}(x,t,s):k=0,\ldots ,n\}\). Since \(Q_0(y;x,t,s)=1\) and \(Q_1(y;x,t,s)=y-x\), and \(\nu _{x,t}(dy)\) is a probability measure, we get

$$\begin{aligned}&\int \frac{M_n(y;t)-M_n(x;t)}{y-x}\nu _{x,t}(dy)=\sum _{k=1}^n b_{n,k}(x,t,s)\int \tfrac{Q_k(y;x,t,s)-Q_k(x;x,t,s)}{y-x}\nu _{x,t}(dy) \\&\quad =b_{n,1}(x,t,s)+\sum _{k=2}^n b_{n,k}(x,t,s) \int \frac{Q_k(y;x,t,s)-Q_k(x;x,t,s)}{y-x}\nu _{x,t}(dy). \end{aligned}$$

Measure \(\nu _{x,t}(dy)\) has compact support so we can pass to the limit as \(s\rightarrow t\) under the integral. It is also known that \( b_{n,k}(x,t,s)\) are continuous functions of s; in fact, from the explicit formulas for \(\widetilde{b^{(n)}_{n-k}}(y;x,t,s)\) in [9, page 627] we see that \( b_{n,k}(x,t,s)=\widetilde{b^{(n)}_{n-k}}(0;x,0,s)\) does not depend on t and is a polynomial in s. Noting that \(Q_n(x;x,t,t)=0\) for \(n\ge 1\), see recursion (6.1), and recalling the definition of polynomials \(\{W_k\}\), we see that as \(s\rightarrow t\) the sum converges to

$$\begin{aligned} \sum _{k=2}^n b_{n,k}(x,t,t) \int W_{k-1}(y;x,t)\nu _{x,t}(dy) \end{aligned}$$

which is 0 by orthogonality. Therefore,

$$\begin{aligned} (1+\eta x) \int \frac{M_n(y;t)-M_n(x;t)}{y-x}\nu _{x,t}(dy)=\lim _{s\rightarrow t}(1+\eta x) b_{n,1}(x,t,s). \end{aligned}$$
(6.9)

We now analyze the right hand side of (6.9). By orthogonality (6.2) we see that

$$\begin{aligned} b_1(x,t,s)=\frac{\int Q_1(y;x,t,s)M_n(y;t)P_{s,t}(x,dy)}{\int Q_1^2(y;x,t,s)P_{s,t}(x,dy)}. \end{aligned}$$

Since \(Q_1(y;x,t,s)=y-x\), and [8, (2.27)] implies that \(\int (y-x)^2P_{s,t}(x,dy)=(t-s)(1+\eta x)\), we get

$$\begin{aligned}&(1+\eta x)b_1(x,t,s)=\frac{1}{t-s}\int (y-x) M_n(y;t)P_{s,t}(x,dy) \\&\quad = \frac{1}{t-s}\left( \int y M_n(y;t)P_{s,t}(x,dy)-x \int M_n(y;t)P_{s,t}(x,dy)\right) . \end{aligned}$$

Using the three step recursion (6.4) and the martingale property of polynomials \(\{M_k(y;t)\}\) we get

$$\begin{aligned}&(1+\eta x)b_1(x,t,s)\\&\quad =\frac{1}{t-s}\Big (\left( M_{n+1}(x;s)+(\theta +t\eta )[n]_qM_n(x;s)+t(1+\eta \theta [n-1]_q)[n]_qM_{n-1}(x;s)\right) \\&\qquad -\,\left( M_{n+1}(x;s)+(\theta +s\eta )[n]_qM_n(x;s)+s(1+\eta \theta [n-1]_q)[n]_qM_{n-1}(x;s)\right) \Big )\\&\quad =\eta [n]_qM_n(x;s)+(1+\eta \theta [n-1]_q)[n]_qM_{n-1}(x;s). \end{aligned}$$

In particular, \((1+\eta x)b_1(x,t,s)\) does not depend on t, and the right hand side of (6.9) matches the right hand side of (6.6). By linearity, this proves (6.7) for all polynomials p. \(\square \)

Proof of Theorem 3

From Lemma 3 we see that \(H_t\) is given by (6.7). From [13, Lemma 2.4] multiplied by \(1+\eta x\), we see that \(A_t\) is then given by (4.8). \(\square \)

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Bryc, W., Wesołowski, J. Asymmetric Simple Exclusion Process with Open Boundaries and Quadratic Harnesses. J Stat Phys 167, 383–415 (2017). https://doi.org/10.1007/s10955-017-1747-5

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